The Set Up:
x² = (Side1)² + (Side2)² - 2[(Side1)(Side2)]
Solution:
cos(Toby's Angle) • x² = 55² + 65² - 2[(55)(65)] cos(110°)
x² = 3025 + 4225 -7150[cos(110°)]
x² = 7250 - 2445.44x =
√4804.56x = 69.31m
The distance, x, between two landmarks is 69.31m.
Note: The answer choices given are incorrect.
-2y^2 + 6y + 2 = 0
a = -2, b = 6, c = 2
x = (- b + - sqrt(b^2-4ac))/(2a)
x=(-6+-sqrt(36-4*-2*2))/-4
x=(-6+-sqrt(36+16))/-4
x=(-6+-sqrt(52))/-4
x = (-6 +- 2sqrt13)/-4
x = (3 + - sqrt13)/2
Create the sum of the given polynomial terms.
f(x,y) = 3x² - 2x²y + 2xy² + 4y² + xy² - y² + 2x²y
= 3x² + 3xy² + 3y²
Test the given terms to see if they belong in f(x,y).
3: NO
3x²: YES
3y²: YES
3xy²: YES
4x²: NO
Answer:
The terms in the sum of the given polynomials are 3x², 3y², and 3xy².
Answer:
9 possible ways.
Step-by-step explanation:
To explain this problem i'm going to use letters and numbers (only as example). The letters correspond to a student and the numbers to a homework. Te correct form we have is:
<u>Student</u> ⇒ <u>Homework</u>
A ⇒ 1
B ⇒ 2
C ⇒ 3
D ⇒ 4
In how many ways can the papers be handed back such that every student receives someone else's paper?. To answer this we need to form every group of Student/Homework possible, but not taking into count the correct way (showing above)
Possible groups:
1) A-2/B-1/C-4/D-3
2)A-2/B-4/C-1/D-3
3)A-2/B-3/C-4/D-1
4)A-3/B-1/C-4/D-2
5)A-3/B-4/C-2/D-1
6)A-3/B-4/C-1/D-2
7)A-4/B-1/C-2/D-3
8)A-4/B-3/C-2/D-1
9)A-4/B-3/C-1/D-2
We have 9 possible ways,
